# Limits of functions with converging zeros

What can one say about the derivatives of a smooth function of several variables that is a limit of smooth functions with converging zeros?

More precisely, suppose that $f_i: R^n \to R^m$ is a sequence of smooth functions converging uniformly in all derivatives to a function $f: R^n \to R^m$. Suppose that $z_{i,k} \in R^n$ are distinct zeroes of $f_i$ for $k = 1,\ldots, l$ converging to $0$ as $i \to \infty$ for all $k$. Clearly some partial derivatives of $f$ have to vanish at $0$, but which ones depends on how the points $z_{i,k}$ converge to $0$.

Let $d(l,n,m)$ be the function such that for any such situation above for $l,n,m$ fixed, there exists a direction $v \in R^n$ for which the first $d(l,n,m)$ directional derivatives of $f$ in the direction of $v$ vanishes at $0$. What is $d(l,n,m)$?

This is related to the question of which partial derivatives at $0$ are approximated by finite difference operators constructed from the points $z_{i,k}$. My question is how this depends on the limiting geometry of $z_{i,k}$, and what statements are independent of the geometry. There must be a literature on this, but I am having trouble finding the right "tag".

-
Take the sequence $f_i(x) = |x|^2-1/i$. This vanishes on infinitely many distinct points near $0$. The limit, $|x|^2$, has just one vanishing partial derivative in each direction. – Will Sawin Feb 27 '14 at 18:45
thanks! Is there a condition on $z_{i,k}$ which guarantees that $f$ has some vanishing higher partial derivatives? – Chris Woodward Feb 27 '14 at 19:41
Not sure, but this might help: en.wikipedia.org/wiki/General_position – Will Sawin Feb 27 '14 at 19:57
@ChrisWoodward $f(x,y)=e^{y}-e^{x}$ is identically zero on line x=y. but no partial derivative(of any order) vanish. – Ali Taghavi Feb 28 '14 at 20:18
and the constant sequence f,f,f.... converges to f. – Ali Taghavi Feb 28 '14 at 20:20

If the functions $f_i$ fit into a smooth family $f_t$ with $f = f_0$ then the family defines a function $F: R^{n+1} \to R^{m+1}$ with local ring isomorphic to that of $f_0$. Hence the existence of $l$ zeroes of $f_t$ converging to $0$ implies that the local ring of $f_0$ has dimension at least $l$. This is like saying that $f_0$ has a bunch of vanishing partial derivatives, but it is a vacuous statement if $n > m$, which is why my original question was a bit silly.